3.105 \(\int \frac{\cos ^2(c+d x)}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=82 \[ \frac{i a}{8 d (a+i a \tan (c+d x))^2}-\frac{i}{8 d (a-i a \tan (c+d x))}+\frac{i}{4 d (a+i a \tan (c+d x))}+\frac{3 x}{8 a} \]

[Out]

(3*x)/(8*a) - (I/8)/(d*(a - I*a*Tan[c + d*x])) + ((I/8)*a)/(d*(a + I*a*Tan[c + d*x])^2) + (I/4)/(d*(a + I*a*Ta
n[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.0682929, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3487, 44, 206} \[ \frac{i a}{8 d (a+i a \tan (c+d x))^2}-\frac{i}{8 d (a-i a \tan (c+d x))}+\frac{i}{4 d (a+i a \tan (c+d x))}+\frac{3 x}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x]),x]

[Out]

(3*x)/(8*a) - (I/8)/(d*(a - I*a*Tan[c + d*x])) + ((I/8)*a)/(d*(a + I*a*Tan[c + d*x])^2) + (I/4)/(d*(a + I*a*Ta
n[c + d*x]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{a+i a \tan (c+d x)} \, dx &=-\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^3} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{8 a^3 (a-x)^2}+\frac{1}{4 a^2 (a+x)^3}+\frac{1}{4 a^3 (a+x)^2}+\frac{3}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i}{8 d (a-i a \tan (c+d x))}+\frac{i a}{8 d (a+i a \tan (c+d x))^2}+\frac{i}{4 d (a+i a \tan (c+d x))}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,i a \tan (c+d x)\right )}{8 d}\\ &=\frac{3 x}{8 a}-\frac{i}{8 d (a-i a \tan (c+d x))}+\frac{i a}{8 d (a+i a \tan (c+d x))^2}+\frac{i}{4 d (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.249609, size = 78, normalized size = 0.95 \[ -\frac{2 \cos (2 (c+d x))-12 d x \tan (c+d x)+6 i \tan (c+d x)+3 i \sin (3 (c+d x)) \sec (c+d x)+12 i d x-7}{32 a d (\tan (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x]),x]

[Out]

-(-7 + (12*I)*d*x + 2*Cos[2*(c + d*x)] + (3*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + (6*I)*Tan[c + d*x] - 12*d*x*Tan
[c + d*x])/(32*a*d*(-I + Tan[c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.082, size = 98, normalized size = 1.2 \begin{align*}{\frac{-{\frac{3\,i}{16}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{ad}}-{\frac{{\frac{i}{8}}}{ad \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{1}{4\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{3\,i}{16}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{ad}}+{\frac{1}{8\,ad \left ( \tan \left ( dx+c \right ) +i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a+I*a*tan(d*x+c)),x)

[Out]

-3/16*I/a/d*ln(tan(d*x+c)-I)-1/8*I/a/d/(tan(d*x+c)-I)^2+1/4/a/d/(tan(d*x+c)-I)+3/16*I/a/d*ln(tan(d*x+c)+I)+1/8
/a/d/(tan(d*x+c)+I)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 1.91309, size = 159, normalized size = 1.94 \begin{align*} \frac{{\left (12 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/32*(12*d*x*e^(4*I*d*x + 4*I*c) - 2*I*e^(6*I*d*x + 6*I*c) + 6*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)
/(a*d)

________________________________________________________________________________________

Sympy [A]  time = 0.494572, size = 153, normalized size = 1.87 \begin{align*} \begin{cases} \frac{\left (- 512 i a^{2} d^{2} e^{8 i c} e^{2 i d x} + 1536 i a^{2} d^{2} e^{4 i c} e^{- 2 i d x} + 256 i a^{2} d^{2} e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{8192 a^{3} d^{3}} & \text{for}\: 8192 a^{3} d^{3} e^{6 i c} \neq 0 \\x \left (\frac{\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 4 i c}}{8 a} - \frac{3}{8 a}\right ) & \text{otherwise} \end{cases} + \frac{3 x}{8 a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a+I*a*tan(d*x+c)),x)

[Out]

Piecewise(((-512*I*a**2*d**2*exp(8*I*c)*exp(2*I*d*x) + 1536*I*a**2*d**2*exp(4*I*c)*exp(-2*I*d*x) + 256*I*a**2*
d**2*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(8192*a**3*d**3), Ne(8192*a**3*d**3*exp(6*I*c), 0)), (x*((exp(6*I*c
) + 3*exp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-4*I*c)/(8*a) - 3/(8*a)), True)) + 3*x/(8*a)

________________________________________________________________________________________

Giac [A]  time = 1.13896, size = 134, normalized size = 1.63 \begin{align*} -\frac{\frac{6 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a} - \frac{6 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a} + \frac{2 \,{\left (3 \, \tan \left (d x + c\right ) + 5 i\right )}}{a{\left (-i \, \tan \left (d x + c\right ) + 1\right )}} + \frac{-9 i \, \tan \left (d x + c\right )^{2} - 26 \, \tan \left (d x + c\right ) + 21 i}{a{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/32*(6*I*log(I*tan(d*x + c) + 1)/a - 6*I*log(I*tan(d*x + c) - 1)/a + 2*(3*tan(d*x + c) + 5*I)/(a*(-I*tan(d*x
+ c) + 1)) + (-9*I*tan(d*x + c)^2 - 26*tan(d*x + c) + 21*I)/(a*(tan(d*x + c) - I)^2))/d